{1 + A} = - ≤ft( { A/2 + A/2} ) is true if

Correct Answer is : 3π2<A<7π2{{3π }}{2} < A < {{7π }}{2}23π<A<27π

Correct Answer is : 3π2<A<7π2{{3π }}{2} < A < {{7π }}{2}23π<A<27π

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UPSC NDA 2 2017 Mathematics Solved Paper

Question : 38 of 120

Marks: +1, -0

\sqrt {1 + \sin A} = - \left( {\sin \frac{A}{2} + \cos \frac{A}{2}} \right)

is true if

Solution:

Concept:

Use the identity

1+\sin A = (\sin\frac{A}{2} + \cos\frac{A}{2})^2

and analyze the sign condition for the absolute value.

Explanation:

Start from

\sqrt{1+\sin A} = -(\sin\frac{A}{2}+\cos\frac{A}{2})

Rewrite LHS:

\sqrt{1+\sin A} = \sqrt{(\sin\frac{A}{2}+\cos\frac{A}{2})^2} = |\sin\frac{A}{2}+\cos\frac{A}{2}|

Thus the equation becomes

|\sin\frac{A}{2}+\cos\frac{A}{2}| = -(\sin\frac{A}{2}+\cos\frac{A}{2})

This holds only when

\sin\frac{A}{2}+\cos\frac{A}{2} \le 0

Express the sum:

\sin\frac{A}{2}+\cos\frac{A}{2} = \sqrt{2}\sin\left(\frac{A}{2}+\frac{\pi}{4}\right)

We require

\sin\left(\frac{A}{2}+\frac{\pi}{4}\right) \le 0

, which implies

\frac{A}{2}+\frac{\pi}{4} \in [\pi, 2\pi]

(mod

2\pi

Solving gives

\frac{A}{2} \in \left[\frac{3\pi}{4}, \frac{7\pi}{4}\right]

\Rightarrow

A \in \left[\frac{3\pi}{2}, \frac{7\pi}{2}\right]

Among the given options, this matches

\frac{3\pi}{2} < A < \frac{7\pi}{2}

Answer:

\frac{3\pi}{2} < A < \frac{7\pi}{2}

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